Answer to Question #243282 in MatLAB for Vamsi reddy

Question

Matrix equations. Using the variables created in 2 and 3, find the values of xMat, yMat and
zMat below. Use matrix operators.
a. xMat = (aVec ∙ bVec) ∙ aMat-
b. yMat = (bVec ∙ aVec), note that this is not the same as (aVec ∙ bVec)
c. zMat = |cMat|(aMat ∙ bMat)1, where is the determinant of cMat , and T again
indicates the transpose (use det).
cM

Answer 1

I have designed the matrix equations for all the aMat, bMat, cMat values using the matrix operator. I have included the comments for each part of the section and the final output of it.

Let me explain you in simpler and step-by-step procedure:-

A)

Step-1:

The given equation is xMat = (aVec * bVec) * aMat2

The initial step is to know about the matrix operator which consist of many different types. The Vector Multiplication operator which is calculated based on the dot operator which acts as an interface between 'a' and 'b'.

Example:-

aMat = a [ \begin{bmatrix} a1 & a2 & \\ a3 & a4 & \\ a5 & an & \end{bmatrix}]

The value of 'a' is [a1 a2 ... an],

bMat = \begin{pmatrix} b1 & \\ b2 & \\ bn & \end{pmatrix}

The value of 'b' is [b1 b2 ... bn],

Consider the value of aMat and bMat, Finally we get the value of a.b = a1b1 + a2b2 +...+ anbn.

B)

Step-2:

The given equation is

yMat = bVec * aVec

The next step is to make use of the matrix multiplication operator which multiplies the Matrix A with the scalar by multiplying each value / element of A in the scalar.

Example:-

Let us assume that A = [4, 7; -2, 6],

So the value in matrix is A = \begin{bmatrix} 4& 7\\ -2 & 6 \end{bmatrix}

Now the value of b is B = A * 3

So the value in matrix is B = \begin{bmatrix} 12 & 21 \\ -6 & 18 \end{bmatrix}

C)

Step-3:

The given equation is

 yMat = bVec * aVec

The next step is to make use of the matrix multiplication operator which multiplies the Matrix A with the scalar by multiplying each value / element of A in the scalar.

Example:-

Let us assume the value of A is \begin{bmatrix} 7 & 6 & 3 \\ 2& 4 & 5\\ 8 & 3& 1 \end{bmatrix}

So the value in matrix is AT = \begin{pmatrix} 7& 2 &8 \\ 6& 4& 3\\ 3& 5& 1 \end{pmatrix}

Finally, The determinant(det) will be using the formulae:-

det\begin{pmatrix} a11 &a12 \\ a21 & a22 \end{pmatrix} = = a11 . a22 - a21 . a12